Let S be the surface $z = x^2 +y^2$ and let $P(1,1,2)$ be a point on the graph of S. What is the angle between the normal line to the surface S at the point P and the tangent line to the surface S at the point P.
Attempt: First, I found normal line at $P$:
$$\nabla f=<2x,2y,-1>,\;\nabla f(1,1,2)=<2,2,-1>$$
Normal line: $L:\begin{cases} x=1+2t \\ y=1+2t \\ z=2-t \end{cases}$
Here it gets unclear to me what to do next. I tried a finding a tangent plane at $(1,1,2)$ and then any point on this tangent plane, so I have a vector tangent to this tangent plane:
Tangent plane: $2x+2y-z=2$. So, what do I do next? hints please.
Best Answer
I guess that by the very definition of tangent line and normal line at a given point both lines form right angle.